First note that
$$0 = (a + b + c)^2 - (a^2 + b^2 + c^2) = 2(ab + bc + ca).$$
Using $a + b = 1 - c$, we then obtain
$$0 = ab + bc + ca = ab + c(a + b) = ab + c(1 - c), \quad \text{ i.e. } \ ab = c(c - 1)$$
and so $abc = c^2(c - 1) \le 0$ because $a^2 + b^2 + c^2 = 1$ implies $c \le 1$. Finally,
$$\begin{align*} 1 &= (a^2 + b^2 + c^2)(a + b + c) \\ &= a^3 + b^3 + c^3 + ab(a + b) + bc(b + c) + ca(c + a) \\ &= a^3 + b^3 + c^3 + ab(1 - c) + bc(1 - a) + ca(1 - b) \\ &= a^3 + b^3 + c^3 - 3abc + ab + bc + ca, \end{align*}$$
hence $a^3 + b^3 + c^3 = 1 + 3abc \le 1$.